Why is it ok to mix up the order of the time series in a moving block bootstrap?

Question

For a time series consisting of a sequence of observations $x_1, x_2, \dots, x_n$, the moving block bootstrap (See here and wikipedia) is implemented thus (emphasis mine):

1. Pick a block length $k$.
2. Create $n - k + 1$ overlapping blocks of length $k$, so that the first block $x_1^k$ consists of the subsequence $(x_1, \dots, x_k)$, the second $x_2^{k+1}$ is the subsequence $(x_2, \dots, x_{k+1})$, etc.
3. Sample $m = \operatorname{round}(\frac{n}{k})$ blocks with replacement.
4. The bootstrapped observations are then obtained by aligning these blocks back to back, in the order they were picked.

Unlike the regular non-parametric bootstrap, in which the sample order does not make a difference, the moving block bootstrap changes the original chronological ordering of the time series. In the extreme case, where the block length $k = 1$, we can obtain the original sequence in reverse order when $(x_n, \dots, x_1)$ is the bootstrap sample.

This seems counter-intuitive to me. Is there an intuitive explanation on why it is ok to mix up the ordering of the blocks?

Answer 1

Note that the overlapping block bootstrap is not the only block-bootstrap method. My comments should apply to the other forms as well.

The basic idea of a block boostrap is to preserve the order - and hence the characteristics like autocorrelation or heteroskedasticity - locally (within the blocks).

It is sensitive to the choice of $k$. You would choose $k$ large enough to capture the important lags (though this isn't the only consideration, there's several tradeoffs come into the choice). So each block provides its own picture of (and so information to estimate) the time-series characteristics like autocorrelation structure.

Certainly it shuffles the order across the blocks, but if $k$ is not too small (and the series is stationary) that effect should be relatively small compared to the information within-blocks. If you had any serial dependence, you would not choose $k=1$.

Answer 2

I believe $k=1$ would not mix up the order.

Assume $K=1$, we pick $N$ samples from $N$ elements, each sample of size $= 1$, lined back to back, with replacement.

if the list is $(1,2,3,4,5,6,7,8,9,10)$

$x_1 = 1$ with replacement, our list is still $(1,2,3,4,5,6,7,8,9,10)$

$x_2$ would then be $= 2$

$x_3 = 3$

and so forth. Lined back to back, would give us $x_1,x_2,x_3, \ldots, x_n$ in this case, to $X_{10}$ which would then be $(1,2,3,4,5,6,7,8,9,10)$

It preserves the order.